1.07 Absolute value inequalities

Represent the inequality \left|x\right| > 2 on a number line.

This inequality represents values of x which are "more than 2 units away from 0." To plot this, we will need to use two regions. Also, note that this inequality doesn't include the endpoints, so we will use unfilled points to show this.

We can give a quick check of the answer by thinking about whether this is an "and"-type inequality or an "or"-type inequality.

In this case, the inequality \left|x\right| > 2 has a solution of x < -2 \text{ or } x > 2. Our solution on the number line has two distinct parts, which matches what we expect for an "or"-type inequality.

Rewrite the solution to \left|x\right| > 2 in interval notation.

We can use either the number line or x < -2 \text{ or } x > 2 to help write this in interval notation.

The x < -2 part can be written as \left(-\infty, -2\right).

The x >2 part can be written as \left(2,\infty\right).

Since this is an "or" inequality we will find the union of these two sets to give: \left(-\infty, -2\right) \cup \left(2,\infty\right)

Rewrite the solution to \left|x\right| > 2 in set notation.

In the previous parts, we determined that the inequality is an "or"-type because it uses the greater than symbol \left(>\right) and there are two distinct sections of the solution.

Since this is an "or" inequality, we will write the answer in set notation as: \{x | x\lt -2 \text{ or } x\gt 2 \}

Solve the inequality for x. Express your solution using interval notation.

In order to solve this inequality, it will be easier to first remove the absolute value by rewriting the inequality as a compound inequality. We can then solve as normal.

So the solutions are "all values of x between \dfrac{3}{2} and 6 (inclusive)". We can express this using interval notation as the interval:\left[\frac{3}{2},\, 6\right]

Represent the solution set on a number line.

The solution set for this inequality consists of a single interval, so it will only need one region on the number line. The endpoints are also included this time, which we represent using filled points.

We can give a quick check of the answer by thinking about whether this is an "and"-type inequality or an "or"-type inequality.

In this case, we found the solution to the inequality \left|\dfrac{4}{3}x - 5\right| \leq 3 to be the compound inequality \dfrac{3}{2} \leq x \leq 6. A compound inequality is a way of writing an "and"-type inequality, which matches the fact that the number line has one region between two endpoints.

Determine whether x=2.5 is a valid solution to the absolute value inequality.

Plot the solution set together with the indicated point to check if the point lies inside or outside the solution set. Or test algebraically by substituting the value into either the original or rearranged inequality and checking if the resulting statement is true.

Graphically:

Consider the plot of the solution set together with the point at x=2.5.

The point x=2.5 lies on a section of the line that in the solution set, indicated by the green interval, and therefore is a valid solution.

Algebraically:

Substitute x=2.5 into the original inequality.

Considering \dfrac{5}{3} \approx 1.67 this statement is true and x=2.5 is a valid solution.

Solve the inequality.

In order to solve this inequality, we need to isolate the absolute value expression and then rewrite the inequality as a compound inequality. We can then solve by applying the properties of inequality.

Now, we can write |x+2|<14 as a compound inequality and solve.

Represent the solution on a number line.

The solution set for this inequality consists of a single interval, so it will only need one region on the number line. Because neither inequality has an "equal to" the endpoints are not included in the solution, which we represent using unfilled points.

Verify the solution graphically.

In part (a), we found the isolated absolute value expression to be \left |x+2 \right |<14. First, we will create a table of values for the absolute value function y=\left|x+2\right|. Then, we will plot the values from the table and look for the x-values of the points where y<14.

To create a table of values for y=\left |x+2 \right |, we can use the solution from part (a) when choosing x-values to substitute. The solution tells us that the x-values should range from x=-16 to x=12. The vertex of the function will be exactly halfway between these values, so it will be at x=\dfrac{-16+12}{2}=-2.

Next, graph the points on the coordinate grid, and look for the section of the graph where the outputs are strictly less than 14.

We can write our solution in interval notation as (-16,\,12) and in set notation as \{x | -16 \lt x \lt 12 \}.

Write and solve an absolute value inequality that represents the scenario.

A margin of error indicates that the percentages for the poll could be off by 4 percentage points in either direction of 76 \%. It may be helpful to first graph possible solutions to the inequality on a number line and then use the number line to help us write an inequality.

Similar to writing and solving absolute value equations, understanding the distance of the possible solutions from the middle or average value will help us write our statement.

Let v= \text{the percentage of people who said they will vote yes}. If the solutions are up to 4 percentage points from 76 \%, an inequality with the possible solutions of 72 \% to 80 \% is |v-76| \leq 4.

By solving the inequality we wrote, we can confirm that our inequality makes sense in context and that its solution matches the number line solution set.

Determine what the viable solutions to the inequality mean in context.

The margin of error would mean that anywhere from 72 \% to 80 \% of people would likely vote yes on the survey.